Basics [Coordinate Systems/Vector Calculus]

Q.1 An electrostatic potential is given by theta=2xsqrt y volts in the rectangular co-ordinate system. The magnitude of the electric field at x = 1 m, y = 1 m is __ V/m. [1992 1M]

Ans:-  sqrt5

Explaination: – Electrostatic potential, theta=2xsqrt y

thereforeoverrightarrow E;=-grad;left(thetaright)=-nablatheta thereforeoverrightarrow E;=-left[{widehat a}_x;fracpartial{partial x};left(2xsqrt y;right)+{widehat a}_y;fracpartial{partial y};left(2xsqrt y;right)right]; overrightarrow E;=;-{widehat a}<em>x;2sqrt y;-{widehat a}</em>{y;}frac x{sqrt y};

Now, at x = 1 m and y = 1 m

overrightarrow E;=;-2{widehat a}_x;-{widehat a}_y therefore left|overrightarrow Eright|;=;sqrt{2^2;+1^2};=sqrt5;V/m

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Q.2 Given a vector field overrightarrow F, the divergence theorem states that [2002 1M]

(a) int_s;overrightarrow F.;doverrightarrow s;=;int_v;nabla.;overrightarrow F;dV

(b) int_s;overrightarrow F.;doverrightarrow s;=;int_v;nablatimes;overrightarrow F;dV

(c) int_s;overrightarrow Ftimes;doverrightarrow s;=;int_v;nabla.;overrightarrow F;dV

(d) int_s;overrightarrow Ftimes;doverrightarrow s;=;int_v;nablatimesoverrightarrow F;dV

Ans:- (a)

Explanation:- The Divergence Theorem is applicable to the closed surface. It is simply remembered as, the surface integration of the vector field over the closed surface can also be calculated by taking the volume integration of the divergence of that vector field.

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Q.3 If overline E is the electric field intensity, nablacdotleft(nablatimesoverline Eright) is equal to [2005]

(a) overline E

(b) left|overline Eright|

(c) null vector

(d) zero

Ans:- (d)

Explanation:- Divergence of a curl field is zero. This can be easily proved by taking any random vector and applying the successive curl and divergence which will lead to 0.

nablacdotleft(nablatimesoverrightarrow Eright);=0

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Q.4 Divergence of the vector field

Vleft(x,y,zright);=;-left(x;cos;xy;+;xright);widehat i;+;left(y;cos;xyright);widehat j;+;left(sin;z^2;+x^2;+y^2right);widehat k is [2007]

(a) 2;z;cos;z^2

(b) sin;xy;+;2z;cos;z^2

(c) x sin xy – cos z

(d) none of these

Ans:- (a)

Explanation:-
Vleft(x,y,zright);=;-left(x;cos;xy;+;xright);widehat i;+;left(y;cos;xyright);widehat j;+;left(sin;z^2;+x^2;+y^2right);widehat k

begin{array}{l}timesleft[sinleft(Z^2right)+left(x^2right)+left(y^2right)right];k\end{array}

Divergence is given as

nablacdot V=frac{partial V_x}{partial x};+frac{partial V_y}{partial y};+frac{partial V_z}{partial z}

= -cos xy + x(sin xy)y + cos xy – y sin(xy)x + 2z cos z²

= begin{array}{l}2z;cos;z^2\end{array}

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Q.5 The direction of vector A is radially outward from the origin, with left|Aright|;=;kr^n;where;r^2=x^2+y^2+z^2 and k is a constant. The value of n for which nablacdot A=0 is [2012]

(a) -2

(b) 2

(c) 1

(d) 0

Ans:- (a)

Explanation :- begin{array}{l}overrightarrow A;=kr^n;{widehat i}_r\end{array}

begin{array}{l}nablacdotoverrightarrow A;=;frac1{r^2};fracpartial{partial r};left(r^2cdot kr^nright)\end{array} begin{array}{l}=;frac1{r^2};cdotfracpartial{partial r};left(kr^{n+2}right)\end{array} begin{array}{l}=;left(n+2right)kcdotfrac{r^{n+1}}{r^2};=left(n+2right)k;r^{n-1}\end{array} begin{array}{l}nablacdotoverrightarrow A;=;0\n+2=0\n=-2\end{array}

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Q.6 The following four vector fields are given in Cartesian coordinate system. The vector field which does not satisfy the property of magnetic flux density is

(a) y^2{widehat a}_x;+;z^2{widehat a}_y;+;x^2{widehat a}_z;

(b) z^2{widehat a}_x;+;x^2{widehat a}_y;+;y^2{widehat a}_z;

(c) x^2{widehat a}_x;+;y^2{widehat a}_y;+;z^2{widehat a}_z;

(d) y^2z^2{widehat a}_x;+;x^2z^2{widehat a}_y;+;x^2y^2{widehat a}_z

Ans :- (c)

Explanation:- We know that the magnetic flux density i.e. B is always in a closed path due to non-existent of the monopole. So according to the Gauss’ Law for magnetic fields we have, nablacdot B=0 (i.e. solenoidal property)

So cheking each option by calculating the divergence when, overrightarrow B=x^2{widehat a}_x;+y^2{widehat a}_y;+z^2;{widehat a}_z

nablacdot B=2x+2y+2z;neq0

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Q.7 A perfectly conducting metal plate is placed in x – y plane in a right handed coordinate system. A charge of +32mathrmpiin_0;sqrt2 Coulomb is placed at co-ordinate (0,0,2). in_0 is the permittivity of free space. Assume widehat i,;widehat j,;widehat k; to be unit vectors along x, y and z axes respectively. At the co-ordinate left(sqrt2,;sqrt{2,};0right) , the electric field vector E (Newtons/Coulomb) will be [2014]

(a) 2sqrt2;widehat k

(b) -2;widehat k

(c) 2;widehat k

(d) -2sqrt2;widehat k

Ans:- (b)

Explanation:-
According to Image Theory, there is an image of the charge at (0,0,-2). Hence we can say there are two charges. Given +32mathrmpiin_0;sqrt2 at (0,0,2) and -32mathrmpiin_0;sqrt2 at (0,0,-2).

Total field is {overrightarrow E}_1+{overrightarrow E}_2

{overrightarrow E}_1=frac{32mathrmpiin_0;sqrt2}{4mathrmpiin_0;left(2+2+4right)^{3/2}};left(sqrt2;{widehat a}_x;+sqrt2{widehat a}_y;-2{widehat a}_zright) {overrightarrow E}_2=frac{-32mathrmpiin_0;sqrt2}{4mathrmpiin_0;left(2+2+4right)^{3/2}};left(sqrt2;{widehat a}_x;+sqrt2{widehat a}_y;+2{widehat a}_zright) overrightarrow E=-2{widehat a}_z;=-2widehat k;

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Q.8 Consider a function overline f;=;frac1{r^2};widehat r, where r is the distance from the origin and ;widehat r is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is [2015]

(a) 0

(b) ;2mathrmpi

(c) ;4mathrmpi

(d) ;Rmathrmpi

Ans:- (c)

Explanation:- overline f=frac1{r^2};cdotwidehat r

from divergence theorem as we know,

int_v;left(nablacdotoverline fright)dv=oint_soverline fcdot doverline s oint_soverline fcdot doverline s;=oint_s;left(frac1{r^2};cdotwidehat rright);times r^2;sin;thetacdot dthetacdot dphicdotwidehat r =oint_ssin;thetacdot dthetacdot dphi;=left[int_0^pisintheta;dthetaright];left[int_0^{2pi}dphiright]=4mathrmpi

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Q.9 In the cylindrical coordinate system, the potential produced by a uniform ring charge is given by f = f(r,z), where f is a continuous function of r and z. Let overrightarrow E be the resulting electric field. Then the magnitude of nablatimesoverrightarrow E [2016]

(a) increases with r

(b) is 0

(c) is 3

(d) decreases with z

Ans:- (b)

Explanation:- We know that Electric field is irrotational.

Hence, nablatimes E=0

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Q. 10 The line integral of the vector field F=5_xzwidehat i+left(3x^2+2yright);widehat j+x^2zwidehat k along a path from (0,0,0) to (1,1,1) parameterized by left(t,;t^2,;tright) is _. [2016]

Ans :- 4.41

int_c;overline F.overline dr =int_c;5x;z;dx+left(3x^2+2yright);dy;+x^2;z;dz x=t,;y=t^2,;z=t,;t=0;to;1

dx = dt; dy = 2tdt, dz = dt

=int_0^1;5t^2dt+left(3t^2+2t^2right)2t;dt+t^3dt =int_0^1;left(5t^2+11t^3right)dt =left[frac{5t^3}3;+frac{11t^4}4;right]_0^1;=frac53;+frac{11}4;=frac{53}{12};=4.41

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Q.11 The figures show diagrammatic representations of vector fields overrightarrow X,;overrightarrow Y,;;and;;overrightarrow Z , respectively. Which one of the following choices is true? [2017]

(a) nablacdotoverrightarrow X;=0,;nablatimesoverrightarrow Y;neq0,;nablatimesoverrightarrow Z;=0

(b) nablacdotoverrightarrow X;neq0,;nablatimesoverrightarrow Y;=0,;nablatimesoverrightarrow Z;neq0

(c) nablacdotoverrightarrow X;neq0,;nablatimesoverrightarrow Y;neq0,;nablatimesoverrightarrow Z;neq0

(d) nablacdotoverrightarrow X;=0,;nablatimesoverrightarrow Y;=0,;nablatimesoverrightarrow Z;=0

Ans:- (c)

Explanation:- overrightarrow X;is;going;away;so;overrightarrownablacdotoverrightarrow Xneq0 .

overrightarrow Y;is;moving;circulator;direction;so;overrightarrownabla;timesoverrightarrow Yneq0 overrightarrow Z;;has;circular;rotation;so;overrightarrownabla;timesoverrightarrow Zneq0

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